Three point charges of $2 \ mC$ each are kept at the vertices of an equilateral triangle of side $50 \ cm$. If the system is supplied energy at the rate of $2 \ kW$,the time taken to move one of the charges to the midpoint of the line joining the other two charges is (in $s$)

Two equal charges of magnitude $Q$ each are placed at a distance $d$ apart. Their electrostatic energy is $E$. $A$ third charge $-Q / 2$ is brought midway between these two charges. The electrostatic energy of the system is now

The work required to be done to assemble four particles,each having a charge $+q$,at the four corners of a square of side $a$ is .......

Three charges $Q$,$(-2q)$ and $(-2q)$ are placed at the vertices of an isosceles right-angled triangle as shown in the figure. The net electrostatic potential energy is zero if $Q$ is equal to

$A$ short electric dipole has a dipole moment of $16 \times 10^{-9} \, Cm$. The electric potential due to the dipole at a point at a distance of $0.6 \, m$ from the centre of the dipole,situated on a line making an angle of $60^{\circ}$ with the dipole axis is $.........V$. (Given: $\frac{1}{4 \pi \epsilon_{0}} = 9 \times 10^{9} \, Nm^{2}/C^{2}$)

The work done in rotating a dipole placed parallel to the electric field through $180^{\circ}$ is $w$. What is the work done in rotating it through $60^{\circ}$? $(\cos 0^{\circ}=1, \cos 60^{\circ}=\frac{1}{2}, \cos 180^{\circ}=-1)$